Analyse long-run behaviour of a Leslie model

Stable profiles and long-run behaviour

Compare each stage with the same stage in a later state. A common multiplicative factor across all non-zero components supports a stable profile, but it does not mean that the counts are fixed; a repeating sequence of distinct states is a cycle

1. Compare like stages Divide a stage count by the count of the same stage in the preceding state \(\displaystyle \frac{(S_{n+1})_i}{(S_n)_i}\)
2. Use a multiplicative ratio A stable profile is tested with ratios, not with equal additive differences \(\displaystyle \frac{21}{14}=\frac{3}{2}\)
3. Check every non-zero component One common factor is supported only when all corresponding non-zero components agree across the required transitions \(\displaystyle \frac{21}{14}=\frac{30}{20}=\frac{45}{30}=\frac{3}{2}\)
4. Separate stability from equilibrium A stable profile can have changing counts, while a repeated sequence of distinct states is a cycle rather than one equilibrium state \(\displaystyle k=\frac{3}{2}>1\Rightarrow S_1\ne S_0\)

Worked example

A Leslie model has the matrix and initial state shown below. Determine the common factor and describe the stage pattern

1. \(\displaystyle S_1=L S_0=\begin{pmatrix}18+6\\\frac{8}{9}\times18\\\frac{2}{3}\times12\end{pmatrix}=\begin{pmatrix}24\\16\\8\end{pmatrix}\)Apply the Leslie matrix once, using fertility in the first row and survival on the subdiagonal
2. \(\displaystyle \frac{24}{18}=\frac{4}{3},\quad\frac{16}{12}=\frac{4}{3},\quad\frac{8}{6}=\frac{4}{3}\)All three corresponding stage ratios in the first transition are equal
3. \(\displaystyle S_2=L S_1=\begin{pmatrix}24+8\\\frac{8}{9}\times24\\\frac{2}{3}\times16\end{pmatrix}=\begin{pmatrix}32\\\frac{64}{3}\\\frac{32}{3}\end{pmatrix}\)Apply the same Leslie matrix to the next state for a second transition
4. \(\displaystyle \frac{32}{24}=\frac{4}{3},\quad\frac{64/3}{16}=\frac{4}{3},\quad\frac{32/3}{8}=\frac{4}{3},\quad k=\frac{4}{3}>1\)The common factor holds in both transitions and is greater than 1, so the profile is stable while the counts increase

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